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How many arrangements of the letters A, C, C, E, N, T includ

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by Max@Math Revolution » Tue Jun 18, 2019 1:43 am
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How many arrangements of the letters A, C, C, E, N, T include at least one letter between the two C’s?

A. 60
B. 120
C. 240
D. 360
E. 720
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by GMATGuruNY » Tue Jun 18, 2019 2:10 am
Max@Math Revolution wrote:How many arrangements of the letters A, C, C, E, N, T include at least one letter between the two C’s?

A. 60
B. 120
C. 240
D. 360
E. 720
Good arrangements = total arrangements - bad arrangements.

Total arrangements:
Number of ways to arrange 6 elements = 6!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical C's:
6!/2! = 360.

Bad arrangements:
In a bad arrangement, the two C's are in adjacent slots.
Let [CC] represent the 2 adjacent C's.
Number of ways to arrange the 5 elements [CC], A, E, N and T = 5! = 120.

Good arrangements:
Total arrangements - bad arrangements = 360-120 = 240.

The correct answer is C.
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by Max@Math Revolution » Wed Jun 19, 2019 11:53 pm
=>

The easiest way to solve this problem is to find the number of arrangements satisfying the complementary condition that there is no letter between the two C’s and subtract this from the total number of arrangements of the letters. The best approach to solving probability or counting problems, including the words, ‘at least’, is to use complementary counting.

The total number of arrangements of the letters A, C, C, E, N and T is (6!)/(2!) = 360.
To count the number of arrangements with no letter between the two C’s, we consider CC to be one letter. Thus, the number of arrangements satisfying the complementary condition is 5! = 120.

Thus, the number of arrangements of the letters A, C, C, E, N, T with at least one letter between the two C’s is 360 – 120 = 240.

Therefore, the answer is C.
Answer: C
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